Read e-book online A Tale of Two Fractals PDF

By A.A. Kirillov

ISBN-10: 081768381X

ISBN-13: 9780817683818

ISBN-10: 0817683828

ISBN-13: 9780817683825

Since Benoit Mandelbrot's pioneering paintings within the past due Seventies, rankings of analysis articles and books were released regarding fractals. regardless of the amount of literature within the box, the final point of theoretical knowing has remained low; such a lot paintings is aimed both at too mainstream an viewers to accomplish any intensity or at too really expert a group to accomplish frequent use. Written by way of celebrated mathematician and educator A.A. Kirillov, A story of 2 Fractals is meant to aid bridge this hole, delivering an unique remedy of fractals that's right now obtainable to novices and sufficiently rigorous for severe mathematicians. The paintings is designed to provide younger, non-specialist mathematicians a great starting place within the concept of fractals, and, within the strategy, to equip them with publicity to numerous geometric, analytical, and algebraic instruments with functions throughout different areas.

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Extra info for A Tale of Two Fractals

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D e 3 , a cube root of 1. ? S / for such a system? 3. Continued Fractions There is one more interesting numerical system related to the notion of continued fraction. Let k D fk1 ; k2 ; : : : g be a finite or infinite system of positive integers. 4) as n ! 1 if the sequence k is infinite. It is well known that the limit in question always exists. 0; 1/ is the value of the unique infinite continued fraction. 0; 1/, they can be values of two different finite continued fractions: k D fk1 ; : : : ; kn 1 ; 1g and k 0 D fk1 ; : : : ; kn 1 C 1g.

K/ ; 2 k+1 2 Œ0; 1, we have . / ˙ 1 1 D k a+ . kC1 2 / 2 . /C3 . 3) . n Corollary. l/ (here we have not only congruence but in fact equality, since in this case, 21 D 1). Proof of the theorem. Consider the triangular piece of the infinite gasket that is based on the segment Œk 1; k C 1. It is shown in Fig. 4. We denote the values of at the points k 1; k; k C1 by a ; a; aC respectively. Then the values bC ; b ; c in the remaining vertices shown in Fig. l/ is an integer when l < 2n . 42 3 Harmonic Functions on the Sierpi´nski Gasket The result is c D 5a 2a 3a C 2aC ; 5 bC D 2a 2aC ; b D 2a 2aC C 3a : 5 Consider now the functions g˙ W !

1 First Properties of Harmonic Functions We start with the following fact. 1. S1 / of all harmonic functions on S1 has dimension 3. S1 / are the values of this function at three boundary points. Proof. From linear algebra, we know that if a homogeneous system of linear equations in n variables has only the trivial solution, then the corresponding inhomogeneous system has a unique solution for any right-hand side. Sn / D 3 for all n 1. Therefore, every harmonic function on Sn has a unique harmonic extension to SnC1 , hence to S1 .

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A Tale of Two Fractals by A.A. Kirillov


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